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Let $K$ be a number field (i.e. a finite extension of $\Bbb{Q}$). Let $\mathfrak{p}$ be a prime ideal of the ring of integers $\mathcal{O}_K$. Also, let us denote the completion of $K$ with respect to $\mathfrak{p}$ (i.e. wrt. the absolute value defined by $\mathfrak{p}$) by $K_\mathfrak{p}$.

Let $L/K_\mathfrak{p}$ be a finite extension and suppose $$ \mathfrak{p} \mathcal{O}_L = \prod_i \mathfrak{p_i}^{m_i} $$ is the factorization of $\mathfrak{p} \mathcal{O}_L$ into prime ideals where $\mathfrak{p}_i \subseteq \mathcal{O}_L$ are prime ideals of $\mathcal{O}_L$.

Question: Is there a $\mathfrak{p}_i$ such that $L = K_{\mathfrak{p}_i}$?

I was interested in this question because in this comment of my other post, it sounded like this is a general statement. That's why I would like to try the theory behind that better.

ANT Learner
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  • But $K_{\mathfrak{p}}$ is a local field, right? And so are all finite extensions of it. – Jeroen van der Meer Jan 18 '21 at 10:39
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    Since $L/K_{\mathfrak{p}}$ is an extension of local fields $\mathcal{O}_L$ is a DVR, so there can only be one such $\mathfrak{p}_i$, so that $\mathfrak{p}$ factors as $\mathfrak{P}^e$ where $\mathfrak{P}$ is the maximal ideal of $\mathcal{O}_L$, and $e$ is the ramification index. – Mummy the turkey Jan 18 '21 at 10:39
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    Moreover, there is no reasonable way of viewing $\mathfrak{p}_i$ as a prime ideal of $K$ - possibly you mean to ask if there is some number field such that $L$ is a localisation of this. – Mummy the turkey Jan 18 '21 at 10:43

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$K_{\mathfrak{p}}$ is a local field, hence so are its finite extensions. So if you have a finite extension $L$, then $\mathfrak{p}$ splits as $\mathfrak{P}^{e}$ for a single lifted prime $\mathfrak{P}$ of $L$, with ramification index $e$. To ask whether $L$ arises as a localisation of $K$ by $\mathfrak{P}$ seems slightly senseless because $\mathfrak{P}$ doesn't live in $K$.

A better way to phrase the question is whether $L$ is still the completion of a global field at a prime, which is indeed the case, as you learned a few days ago. :)

Jeroen van der Meer
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  • I see that my lack of understanding for this topic led to this mistake, thanks for pointing out! Remaning question to be sure: Indeed, in the previous post of mine you cited, there is a global field $F/K$ such that $L$ is the completion of $F$ with respect to some ideal. Does this ideal has to be $\mathfrak{P}$, i.e. $F_\mathfrak{P} = L$? I also have some trouble understanding what you mean by "lifted prime". – ANT Learner Jan 18 '21 at 21:05